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Gravitation Part 1

Section: Physics

Question : 4 of 100

Marks: +1, -0

Solution:

For four equal masses at the corners of a square inscribed in a circle of radius

R

, each mass is attracted by two adjacent masses at distance

\sqrt{2}R

and one opposite mass at distance

2R

. The net inward force is

2\left[\frac{GM^2}{2R^2}\right]\cos 45^\circ + \frac{GM^2}{4R^2} = \frac{GM^2}{R^2}\left(\frac{1}{\sqrt{2}} + \frac{1}{4}\right)

. Equating this to the required centripetal force

Mv^2/R

gives

v^2 = \frac{GM}{R}\left(\frac{1}{\sqrt{2}} + \frac{1}{4}\right) = \frac{GM(2\sqrt{2}+1)}{4R}

, so

v = \frac{1}{2}\sqrt{\frac{GM(2\sqrt{2}+1)}{R}}

. Thus option seq 1 is correct.

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